如何创建一个接受多个参数的函数以在 JavaScript 中执行算术计算？

Question

I want to write a function that performs arithmetic on ALL the arguments supplied. So for instance, if I do:

calculate('+', 3, 5, 6)

It should return 14 (which is 3+5+6)

Or if I do

calculate('*', 6,3,6,8,)

It should return 864 (which is the equivalent of multiplying all those numbers).

The function should essentially be able to handle any amount of numbers I supply to it, while also being able to handle the main arithmetic operators such as + - / *

I'm new at programming. I've tried:

function mCalc(_operator){
  if(_operator=='+'){
    return arguments + arguments;
  }

}
console.log(mCalc('+',5,5));

this is not working so i can't even move forward.

Answer 1

In each function you have an arguments object see the section Rest, default, and destructured parameters as it states:

The arguments object can be used in conjunction with rest, default, and destructured parameters.

function foo(...args) { return args; }

Once you have all the arguments which are needed for your calculation just use Array.prototype.reduce() . As the documentation which states:

The reduce() method executes a reducer function (that you provide) on each element of the array, resulting in a single output value.

I guess you can use as the following:

 const mCalc = (_operator, ...args) => { if(_operator === '+') { return args.reduce((a, c) => a + c, 0); } // rest what you want to implement } const result = mCalc('+', 3, 5, 6, 2); console.log(result);

I hope that helps!

Answer 2

You could take an object for the operators and reduce the values by using a function which returns a function for two operands.

By calling the function the operator is taken and the values are taken to an array by using rest parameters ... .

This approach uses arrow functions , like

calculate = (op, ...values) => values.reduce(take(op));
^^^^^^^^^                                               name of the function/variable
            ^^^^^^^^^^^^^^^                             parameters
                            ^^                          arrow
                               ^^^^^^^^^^^^^^^^^^^^^^^  return value

 const operators = { '+': (a, b) => a + b, '*': (a, b) => a * b }, take = op => operators[op], calculate = (op, ...values) => values.reduce(take(op)); console.log(calculate('+', 3, 5, 6)); // 14 console.log(calculate('*', 6, 3, 6, 8));